WHAT IS THE AREA OF AN N-SIDED IRREGULAR POLYGON?

WHAT IS THE AREA OF AN N-SIDED IRREGULAR POLYGON?

The typical way to measure the area of a piece of land with straight-line boundaries is to note the 2D coordinates [xn,yn] of the corners connecting neighboring lines. Thus if we have an irregular three sided area having corners at [0,0], [2,3] and [1,4], the side-lengths are L1=sqrt(13), L2=sqrt(2), and L3=sqrt(17). One can then use the Heron formula to get the area-

A

s(s L1)(s L2 )(s L3 )

where

s

( L1

L2 2

L3 )

is the semi perimeter

An alternate, and much easier, way to get the area is to take half the absolute value of the vector product of the vectors representing two of the sides. Thus-

i jk

A

1 2

abs 2 1

3 4

0 0

5 2

This second approach of finding the area can be extended to any N sided irregular polygon which can always be broken up into N triangles. Then adding up the sub-areas An produces the total area.

Consider the following schematic of an N sided irregular polygon containing the coordinate origin (0,0) lying at some point within it-

The sub-triangle shaded in gray has the area-

i jk

An

1 2

abs[( xni

yn

j)

x

( xn1i

yn1

j)]

1 2 abs

xn

xn1

yn yn1

0 0

1 2

abs(

xn

y

n1

xn1 yn )

Note that An also represents half the area of a rhombus having two of its sides have length ?

Ln xn 2 yn 2

and

Ln1

x2 n 1

y2 n 1

Next, adding all N triangles making up the polygon produces the area-

A

1 2

N

abs[ xn yn1

n 1

xn1 yn ]

This shows we only need the coordinates of each of the N corners of the polygon to find its total area. It should produce correct values for both convex polygons such as a hexagon or for concave polygons such as stars.

Let us begin by asking for the largest area an equilateral triangle can have and still fit into a circle of radius R. Clearly the three vertexes of this triangle must lie on the circle at 120? intervals. So one can choose the vertex coordinates, expressed in Cartesian coordinates, to be ?

(R, 0), (-R/2, Rsqrt(3)/2), and (-R/2, -Rsqrt(3)/2)

The area will thus be given by-

i

j

k

A

1 2

abs

3R 3R

/2 /2

R 3/2 R 3/2

0 0

33 4

R2

The number 3sqrt(3)/4=1.299078.. is a little surprising since it says the equilateral triangle fills only 1.2990../=0.41349..of the circle area which is given by R2.

Consider next the area of a hexagon of side-length s=1. Here the total area is given by six equal isosceles sub-triangles where the vertex coordinates of one of these can be-

(0, 0), (1,0) and (1/2, sqrt(3)/2)

The total area of the regular hexagon then is-

i

A

6{12

abs

1 1/ 2

jk

0 3/2

0} 0

33 2

Circumscribing the hexagon with a circle of unit radius shows that the hexagon fills 3sqrt(3)/(2) or about 83% of the circle.

As already stated earlier the present area determination method will also work when part of the polygon boundary is concave. This will be the case for stars, Consider the area of the following five pointed star known as the pentagram-

We show one of the ten equal sub-triangles which make up the pentagram in gray This sub-triangle has vertexes at ?

(0, 0), (L1, 0), and The pentagram area will thus be-

( L3cos(/5), L3sin(/5))

i

A

10{12

abs

L3

L1 cos(

/

5)

j 0 L3 sin( / 5)

k

0 0

}

5L1

L3

sin(5

)

Looking at the pentagram geometry, using the law of sines, and having the sidelength of the surrounding pentagon be s=1, we find-

L 1

1 2 sin(

/

5)

,

L2

1 2 cos(

/ 5)

and

L3

sin( /10) sin(2 / 5)

Thus the area of a pentagram inscribed by a pentagon of side-length s=1 is-

A

(

5 2

)(

sin( /10) sin(2 / 5)

8 cos(

5 /10) cos(

/ 5)

0.8122992405...

We can check this answer by noting that the pentagram area just equals the area Ap=(5/4)(1/tan(/5)) of the circumscribing pentagon minus five times the area of the isosceles triangle whose sides have lengths L2, L2, and 1. The area of each of these sub-triangles is An=(1/4)tan(3/10). The area of the pentagram thus becomes-

A

(

5 4

)

1

tan(

/

5)

1 tan(3

/ 10)

0.81229924...

which checks with the earlier result. An interesting sideline of the pentagram is that the oblique straight line distance from one of its vertexes to one two away equals exactly-

sin(3 / 5) sin( / 5)

3 4 sin2 (

/ 5)

1

2

5

1.680339...

You will recognize that is just the Golden Ratio already well known to the ancient Greeks.

Let us end the discussion by looking at the area of the four sided irregular polygon having corners at (3,0), (-1,2), (0,0), and(-2,-3) as shown-

There are two ways to evaluate the area of this polygon. One can add together the areas defined by the sub-triangles ABC and DAC or one can subtract area DCB outside the polygon from the large triangle DAB. Using the second approach we find-

A

1 2

abs

i 4 5

j 2 3

k

i

0 abs 1

0

2

j 2 3

k

0

7.5

0

The same result will be produced by taking the other route.

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